removing double elements in an array - c++

Question

I am trying to remove double elements in an array. I developed a simple code, but it is still not working. Is it possible to hint for some input maybe I haven't tried. I tried corner and test cases. The following is the problem statement:
A sequence of numbers given. Remove element’s doubles, leaving first copy.
Input: Contains a natural n (n ≤ 100000) – the n quantity numbers in a sequence, then n non-negative numbers – elements of the sequence which module is not greater than 999.
output: changed sequence.
It seems I can't get what might be the problem
#include <iostream>
//#include <cmath>
//#include <climits>
#define SIZE 100000
using namespace std;
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int n, k, p;
bool tag; tag = false;
cin >> n;
long long int *a = new long long int[n];
long long int b[SIZE];
for (int i = 0; i < n; i++) { cin >> a[i]; }
for (int i = 0; i < n; i++) { k = 0;
for (int j = i + 1; j < n; j++) {
if (a[i] == a[j]) { b[k] = j-k; k++; tag = true; }
}
if (tag) {
for (int i = 0; i < k; i++) {
p = b[i];
for (int i = p; i < n; i++) { a[i] = a[i + 1]; }
n--;
}
tag = false;
}
}
for (int i = 0; i < n; i++) { cout << a[i] << " "; }
return 0;
}
Input: 6 1 2 2 4 3 4 Output: 1 2 4 3

Answer 1

You can use unordered_set and vector
int n; cin >> n;
long long int x;
unordered_set<long long int>myset;
vector<long long int>v1;
for (int i = 0; i < n; i++)
{
cin>>x;
if(myset.find(x)==myset.end())
{
myset.insert(x);
v1.push_back(x);
}
}
for(int i=0;i<v1.size();i++)
{
cout<<v1[i]<<" ";
}

Answer 2

You could use in you advantage the fact that input values are in the range from 0 to 999.
A simple bool used[1000]{} could be used to flag if the current value has been used already before pushing it to cout, thus ensuring both O(n) complexity and limited memory usage (1000 bytes for the bool[]}.
Here's a sample solution around this idea:
#include<iostream>
#define MAX_VALUE 999
using namespace std;
int main() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
bool used[MAX_VALUE + 1]{};
size_t n;
cin >> n;
for (size_t num, i = 0; i < n; ++i) {
cin >> num;
if (!used[num]) {
cout << num << " ";
used[num] = true;
}
}
return 0;
}

Answer 3

You could try creating a second array of unique numbers as you go. I will demonstrate with a vector for the sake of simplicity.
std::vector<int> v;
for (int i = 0; i < n; i++) {
if (std::find(v.begin(), v.end(), arr[i]) == v.end()) {
v.push_back(arr[i]);
}
}
Then, you just write the contents of the vector to the output file.

Answer 4

Here is my version of O(n) complexity. Your solution may exceed time-limit ( if it is low )
bool check[2000];
for (int i = 0; i < 2000; i++) check[i] = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
// +999 to avoid negative numbers
check[a[i] + 999] = 1;
}
bool isPrint = false;
for (int i = 0; i < n; i++) {
if (check[a[i] + 999]) {
// mark false if already printed
check[a[i] + 999] = 0;
if (isPrint) printf(" ");
printf("%d", a[i]);
isPrint = true;
}
}